Find the minimum value of
\[\frac{x^2}{y - 1} + \frac{y^2}{x - 1}\]for real numbers $x > 1$ and $y > 1.$
Solution: Let $a = x - 1$ and $b = y - 1.$  Then $x = a + 1$ and $y = b + 1,$ so
\begin{align*}
\frac{x^2}{y - 1} + \frac{y^2}{x - 1} &= \frac{(a + 1)^2}{b} + \frac{(b + 1)^2}{a} \\
&= \frac{a^2 + 2a + 1}{b} + \frac{b^2 + 2b + 1}{a} \\
&= 2 \left( \frac{a}{b} + \frac{b}{a} \right) + \frac{a^2}{b} + \frac{1}{b} + \frac{b^2}{a} + \frac{1}{a}.
\end{align*}By AM-GM,
\[\frac{a}{b} + \frac{b}{a} \ge 2 \sqrt{\frac{a}{b} \cdot \frac{b}{a}} = 2\]and
\[\frac{a^2}{b} + \frac{1}{b} + \frac{b^2}{a} + \frac{1}{a} \ge 4 \sqrt[4]{\frac{a^2}{b} \cdot \frac{1}{b} \cdot \frac{b^2}{a} \cdot \frac{1}{a}} = 4,\]so
\[2 \left( \frac{a}{b} + \frac{b}{a} \right) + \frac{a^2}{b} + \frac{1}{b} + \frac{b^2}{a} + \frac{1}{a} \ge 2 \cdot 2 + 4 = 8.\]Equality occurs when $a = b = 1,$ or $x = y = 2,$ so the minimum value is $\boxed{8}.$